a less than a greater; which is abfurd. Therefore F G is not perpendicular to D E. And in the fame manner we prove, that no other Right Line but F C is perpendicular to DE. Wherefore F C is perpendicular to D E. Therefore, if any Right Line touches a Circle, and from the Centre to the Point of Contact a Right Line be drawn, that Line will be perpendicular to the Tangent; which was to be demonftrated. PROPOSITION XIX. THEOREM. If any Right Line touches a Circle, and from the Point of Contact a Right Line be drawn at Right Angles to the Tangent, the Centre of the Circle fhall be in the faid Line. LET any Right Line D' E touch the Circle A B C in C, and let C A be drawn from the Point C at Right Angles to DE. I fay, the Circle's Centre is in A C. For, if it be not, let F be the Centre, if poffible; and join C F. Then, because the Right Line D E touches the Circle A BC, and FC is drawn from the Centre to the Point of Contact; F C will be perpendicular to DE*. And fo the Angle FCE is a Right one. 18 of /bis But ACE is alfo a Right Angle : Therefore the From the Angle FCE is equal to the Angle ACE, a lefs to a Hyp. greater; which is abfurd. Therefore F is not the Centre of the Circle A B C. After this manner we prove, that the Centre of the Circle can be in no other Line, but A C. Wherefore, if any Right Line touches a Circle, and from the Point of Contact a Right Line be drawn at Right Angles to the Tangent, the Centre of the Circle fhall be in the faid Line; which was to be demonftrated. 5.1. † 32. I. PROPOSITION XX. THEOREM. The Angle at the Centre of the Circle is double to the Angle at the Circumference, when the fame Arc is the Bafe of both Angles. LETABC be a Circle, at the Centre E whereof is the Angle BEC, and at the Circumference, the Angle B AC, both of which ftand upon the fame Arc BC. 1 fay, the Angle BEC is double to the Angle B AC. For join A, and produce it to F. Then, because EA is equal to E B, the Angle E A B fhall be equal to the Angle E B A*. Therefore the Angles EA B, E B A, are double to the Angle EAB; but the Angle BEF is equal to the Angles E A B, EBA; therefore the Angle B E F is double to the Angle EAB. For the fame Reason, the Angle FEC is double to EAC. Therefore the whole Angle BEC is double to the whole Angle BAC. Again, let there be another Angle B DC; and join D E, which produce to G. We demonftrate in the fame manner that the Angle G E C is double to the Angle GDC; whereof the Part GEB is double to the Part G D B. And therefore the remaining Part B E C is double to the remaining Part BDC. Confequently, An Angle at the Centre of a Circle is double to the Angle at the Circumference, when the fame Arc is the Bafe of both Angles; which was to be demonftrated. PROPOSITION XXI. Angles that are in the fame Segment of a Circle, LET ABCDE be a Circle, and let BAD, BAED. Lay, those Angles are equal. For, For, let F be the Centre of the Circle A B C DE; and join B F, F D. Now, because the Angle B F D is at the Centre,' and the Angle BAD at the Circumference, and they ftand upon the fame Arc BCD; the Angle BFD will be double to the Angle BAD. For the fame 1 20 of this. $ Reafon, the Angle B FD is alfo double to the Angle BED. Therefore the Angle B AD will be equal to the Angle BE D. If the Angles BAD, BED, are in a Segment lefs than a Semicircle, let A E be drawn; and then all the Angles of the Triangle A B G are + equal to † 32. 1. all the Angles of the Triangle D E G. But the Angles A BE, ADE, are equal, from what has been betore proved; and the Angles AGB, DGE, are alfo equal; for they are vertical Angles. Where- 15. 1. fore the remaining Angle B A G is equal to the remaining Angle GED. Therefore, Angles that are in the fame Segment of a Circle, are equal to each other; which was to be demonftrated. PROPOSITION XXII. The oppofite Angles of any quadrilateral Figure, LET ABDC be a Circle, wherein is defcribed the quadrilateral Figure ABCD. I fay, two oppofite Angles thereof are equal to two Right Angles. For join AD, B C. Then, because the three Angles of any Triangle are equal to two Right Angles, the three Angles of the ⚫ 32. 1. Triangle A B C, viz. the Angles CA B, A B C, BCA, are equal to two Right Angles. But the An gle A B C is † equal to the Angle ADC; for they † 21 of tbis. are both in the fame Segment ABD C. And the Angle ACB is + equal to the Angle A D B, becaufe they are in the fame Segment ACDB; therefore the whole Angle BDC is equal to the Angles A B C, ACB; and if the tommon Angle BAC be added, then the Angles BAC, ABC, AC B, are equal to the Angles BA C, BDC; but the Angles B A C, ABC, G 3 § 32. I. Dif. 11. of this. † 16. 1. ABC, ACB, are equal to two Right Angles. Therefore likewife, the Angles BAC, BDC, fhall be equal to two Right Angles. And after the fame Way we prove, that the Angles A B D, ACD, are alfo equal to two Right Angles. Therefore, the oppofite Angles of any quadrilateral Figure, defcribed in the Circle, are equal to two Right Angles; which was to be demonftrated, PROPOSITION XXIII. THEOREM. Two fimilar and unequal Segments of two Circles cannot be fet upon the fame Right Line, and on the fame Side thereof. OR if this be poffible, let the two fimilar and unequal Segments A C B, A D B, of two Circles, ftand upon the Right Line A B on the fame Side thereof. Draw A CD, and let C B, BD, be joined. Now, because the Segment ACB is fimilar to the Segment A D B, and fimilar Segments of Circles are* fuch which include equal Angles; the Angle A CB will be equal to the Angle A D B; the outward one to the inward one; which is abfurd. Therefore, fimilar and unequal Segments of two Circles, cannot be fe upon the fame Right Line, and on the fame Side thereof; which was to be demonftrated. SITION PROPOSITION XXIV. THEOREM. Similar Segments of Circles, being upon equal L ET AEB, CF D, be fimilar Segments of Circles, ftanding upon the equal Right Lines AB, CD. I fay, the Segment A E B is equal to the Segment C F D. For the Segment A E B being applied to the Segment C F D, fo that the Point A coincides with C, and the Line A B with CD; then the Point B will coincide with the Point D, fince A B and C D are equal. equal. And fince the Right Line A B coincides with CD, the Segment A E B will coincide with the Segment CF D. For if at the fame time that A B coincides with C D, the Segment A E B fhould not coincide with the Segment C F D, but be otherwise, as CGD; then a Circle would cut a Circle in more Points than two, viz. in the Points C, G, D; which is impoffible. Wherefore, if the Right Line AB 10 of this, * * coincides with C D, the Segment A E B will coincide with, and be equal to, the Segment C F D. Therefore, fimilar Segments of Circles, being upon equal Right Lines, are equal to one another; which was to be demonftrated, PROPOSITION XXV. A Segment of a Circle being given, to defcribe the LET ABC be the Segment of a Circle given. It * Bifect A C in D, and let D B be drawn + from 10. 1. the Point D at Right Angles to AC; and join AB. † 11. 1. Now the Angle ABD is either, greater, equal, or lefs, than the Angle BAD. And first let it be greater, and make the Angle BAE at the given Point A, † 23. 1. with the Right Line B A, equal to the Angle ABD; produce BD to E, and join E C. Then, because the Angle ABE is equal to the Angle BA E, the Right Line B E will be equal to * 6. x. E A. And because A D is equal to DC, and DE common, the two Sides A D, D E, are each equal to the two Sides CD, DE; and the Angle A D E is equal to the Angle CDE; for each is a Right one. Therefore the Base A E is equal to the Bafe E C. But AE has been proved to be equal to E B. Wherefore BE is alfo equal to E C. And accordingly the three Right Lines A E, EB, E C, are equal to each other. Therefore a Circle described about the Centre E, with either of the Distances A E, E B, E C, + fhall pass + 9 of this. thro' the other Points, and be that required to be defcribed. But it is manifeft, that the Segment ABC manifeft; is |